The half-life period of a first order chemical reaction is 6.93 min. the time required for the completion of 99% of the chemical reaction will be (log 2=0.302)

Question

Dr. Abubacker · Accepted Answer

(c)
Half-life=6.93 min
$k = \frac{0.693}{6.93} = 0.1$
We know ,$k_{1}$for per cent completion
$k_{1} = \frac{2.303}{t}\log\left( \frac{100}{1} ight)$
$0.1 = \frac{2.303}{t}\log\frac{100}{1}$
$0.1 = \frac{2.303}{t}\log 10^{2}$
$t = \frac{2.303 	imes 2}{0.1} = 46.06$

Q. The half-life period of a first order chemical reaction is 6.93 min. the time required for the completion of 99% of the chemical reaction will be (log 2=0.302)

Explanation

Reviewed by Dr. Abubacker

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